Query Questions 31

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course Phy 241

November 20 around 10:30am.

031. `query 31 *********************************************

Question: `qUniv. 10.52 (10.44 10th edition). 55 kg wheel .52 m diam ax pressed into wheel 160 N normal force mu =.60. 6.5 m N friction torque; crank handle .5 m long; bring to 120 rev/min in 9 sec; torque required? Force to maintain 120 rev/min? How long to coast to rest if ax removed?

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Your solution:

Final angular velocity: 120*(1min/60sec)*(2pi/1rev)=12.60rad/sec

Angular acceleration: 12.60/9=1.40rad/s^2

Moment of I: 0.5*(55)*(0.52^2)=7.44

Torque=(moment of I)*(angular acceleration)

=7.44*1.40

=10.42mN

I don’t know where to go from here, I know friction plays a part, but I don’t know how to set it up.

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Given Solution:

`a** The system is brought from rest to a final angular velocity of 120 rev/min * 1min/60 sec * 2`pi/1 rev = 12.6 rad/s.

The angular acceleration is therefore

alpha = change in omega / change in t = 12.6 rad/s / (9 sec) = 1.4 rad/s^2, approx..

The wheel has moment of inertia I = .5 m r^2 = .5 * 55 kg * (.52 m)^2 = 7.5 kg m^2, approx..

To achieve the necessary angular acceleration we have

tauNet = I * alpha = 7.5 kg m^2 * 1.4 rad/s^2 = 10.5 m N.

The frictional force between ax and wheel is .60 * 160 N = 96 N at the rim of the wheel, resulting in torque

tauFrictAx = -96 N * .52 m = -50 m N.

The frictional torque of the wheel is in the direction opposite motion and is therefore

tauFrict = -6.5 m N.

The net torque is the sum of the torques exerted by the crank and friction:

tauNet = tauFrictAx + tauFrict + tauCrank so that the torque necessary from the crank is

tauCrank = tauNet - tauFrict - tauCrank = 10.5 m N - (-50 m N) - (-6.5 m N) = 67 m N.

The crank is .5 m long; the force necessary to achive the 60.5 m N torque is therefore

F = tau / x = 67 m N / (.5 m) = 134 N.

If the ax is removed then the net torque is just the frictional torque -6.5 m N so angular acceleration is

alpha = -6.5 m N / (7.5 kg m^2) = -.84 rad/s^2 approx.

Starting at 120 rpm = 12.6 rad/s the time to come to rest will be

`dt = `dOmega / alpha = -12.6 rad/s / (-.84 rad/s^2) = 14.5 sec, approx.. **

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Self-critique (if necessary):

OK, now I see. You have to consider Ffric, frictional torque, net torque and then find angular acceleration according to that. GOT IT!

Good.

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self-critique rating #$&*:

Looks very good.

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